Answer to Question #91233 in General Chemistry for bob

Question

How many moles of H3O+ or OH− must you add to 67.9 mL of strong base solution to adjust its pH from 9.950 to 7.600? Assume a negligible volume change.

Answer 1

1) Due to the fact that thе pH value decreases (shifts to the acidic region), it is necessary to add Н₃О⁺ ions

2)

$$pH = -lgC(Н_3О^+)$$

сonsequently when pH = 9.950:

с(H₃O⁺) = 10^-lgC(H3O+) = 10^-9.550 = 1.122*10^-10 (mol/L)

When pH = 7.600:

c(H₃O⁺) = 10^-7.600 = 2.512*10^-8 (mol/L)

3) Now let's calculate the difference in the number of ions H₃O⁺ that must be added to the solution:

Δс(H₃O⁺) = 2.512*10^-8 - 1.122*10^-10 = 2.501*10^-8 (mol/L)

4) If the volume of solution 67.9 mL (0.0679 L):

$$n = cV$$

n = 2.501*10^-8*6.79*10^-2 = 16.982*10^-10 = 1.698*10^-9 (mol)

Answer: 1.698*10^-9 mol